3.10.70 \(\int \frac {-1-8 e^2 x+8 x^2}{e^2-x} \, dx\)

Optimal. Leaf size=23 \[ \log \left (-\frac {1}{2} e^{-4 x^2} \left (1-\frac {e^2}{x}\right ) x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 0.61, number of steps used = 2, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {698} \begin {gather*} \log \left (e^2-x\right )-4 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - 8*E^2*x + 8*x^2)/(E^2 - x),x]

[Out]

-4*x^2 + Log[E^2 - x]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-8 x+\frac {1}{-e^2+x}\right ) \, dx\\ &=-4 x^2+\log \left (e^2-x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.83 \begin {gather*} 4 e^4-4 x^2+\log \left (e^2-x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 8*E^2*x + 8*x^2)/(E^2 - x),x]

[Out]

4*E^4 - 4*x^2 + Log[E^2 - x]

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fricas [A]  time = 0.69, size = 13, normalized size = 0.57 \begin {gather*} -4 \, x^{2} + \log \left (x - e^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(2)*x+8*x^2-1)/(exp(2)-x),x, algorithm="fricas")

[Out]

-4*x^2 + log(x - e^2)

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giac [A]  time = 0.33, size = 14, normalized size = 0.61 \begin {gather*} -4 \, x^{2} + \log \left ({\left | x - e^{2} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(2)*x+8*x^2-1)/(exp(2)-x),x, algorithm="giac")

[Out]

-4*x^2 + log(abs(x - e^2))

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maple [A]  time = 0.20, size = 14, normalized size = 0.61

method	result	size
default	\(-4 x^{2}+\ln \left (x -{\mathrm e}^{2}\right )\)	\(14\)
norman	\(-4 x^{2}+\ln \left ({\mathrm e}^{2}-x \right )\)	\(14\)
risch	\(-4 x^{2}+\ln \left (x -{\mathrm e}^{2}\right )\)	\(14\)
meijerg	\(-8 \,{\mathrm e}^{4} \left (-x \,{\mathrm e}^{-2}-\ln \left (1-x \,{\mathrm e}^{-2}\right )\right )-8 \,{\mathrm e}^{4} \left (\frac {x \,{\mathrm e}^{-2} \left (3 x \,{\mathrm e}^{-2}+6\right )}{6}+\ln \left (1-x \,{\mathrm e}^{-2}\right )\right )+\ln \left (1-x \,{\mathrm e}^{-2}\right )\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*exp(2)*x+8*x^2-1)/(exp(2)-x),x,method=_RETURNVERBOSE)

[Out]

-4*x^2+ln(x-exp(2))

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maxima [A]  time = 0.34, size = 13, normalized size = 0.57 \begin {gather*} -4 \, x^{2} + \log \left (x - e^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(2)*x+8*x^2-1)/(exp(2)-x),x, algorithm="maxima")

[Out]

-4*x^2 + log(x - e^2)

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mupad [B]  time = 0.06, size = 13, normalized size = 0.57 \begin {gather*} \ln \left (x-{\mathrm {e}}^2\right )-4\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x*exp(2) - 8*x^2 + 1)/(x - exp(2)),x)

[Out]

log(x - exp(2)) - 4*x^2

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sympy [A]  time = 0.09, size = 10, normalized size = 0.43 \begin {gather*} - 4 x^{2} + \log {\left (x - e^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(2)*x+8*x**2-1)/(exp(2)-x),x)

[Out]

-4*x**2 + log(x - exp(2))

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